The pair of lanthanides with the highest third ionization energy is

$\mathrm{Eu}=\left[\mathrm{Xe}\right]4{\mathrm{f}}^{7}5{\mathrm{d}}^{0}6{\mathrm{s}}^2\text{\hspace{0.33em}and\hspace{0.33em}}\mathrm{Yb}=\left[\mathrm{Xe}\right]4{\mathrm{f}}^{14}5{\mathrm{d}}^{0}6{\mathrm{s}}^2$. Removal of 3rd electron requires more energy from these elements.